Class 12th

Radius of circle S touching x-axis and centre  $\left(\alpha ,\beta \right)$ $$ is |. According to given conditions

${\alpha }^2+{\left(\beta -1\right)}^2={\left(\left|\beta \right|+1\right)}^2$

${\alpha }^2=4\beta as\beta >0$

$$ $\therefore$  Required locus is L : x2 = 4y

The area of shaded region = $2{\displaystyle {\int }_0^{4}2\sqrt[]{y}dy}$ ${\displaystyle {}_{}^{}\text{exception 25:}}$

= $4.{\left[\frac{\frac{y^{\frac{3}{2}}}{3}}{2}\right]}_0^4$ ${\frac{\frac{{}^{\frac{}{}}}{}}{}}_{}^{}$

= $\frac{64}{3}$ $\frac{}{}$ square units.

$\Delta v={\displaystyle \int \overrightarrow{E}.d\overrightarrow{x}}$  

$\Delta v=\frac{\sigma }{\in o}\frac{d}{2}$

$C_{New}=\frac{Q}{\Delta v}=\frac{\sigma A2\in o}{\sigma d}=\frac{2\in oA}{d}$

$C_{New}=2{\begin{array}{l}C\\ original\\ \end{array}}_{}$

$\frac{C_{New}}{C_{Original}}=2:1$

$\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$                

$y\frac{dy}{dx}=\frac{y^2-x}{5x+y^2}$            

Let y² = t

$\frac{1}{2}.\frac{dt}{dx}=\frac{t-x}{5x=t}$  

Now substitute, t = yx

$\frac{dt}{dx}=v+x\frac{dv}{dx}$

  $\left|\frac{{\left(v+1\right)}^4}{{\left(v+2\right)}^3}\right|=\frac{C}{x}$

$\left|{\left(y^2+x\right)}^4\right|=C\left|{\left(y^2+2x\right)}^3\right|$    

$\frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}=e^{2x}-\frac{6e^{-x}}{2+9e^{-2x}}$

${\displaystyle \int dy={\displaystyle \int e^{2x}dx-3{\displaystyle \int \frac{e^{-z}}{\underset{put{e}^{-x}=t}{\underset{?}{1+\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)}}}}dx}}$

= $y=\frac{e^{2x}}{2}=\sqrt[]{2}ta{n}^{-1}\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)+C$

It is given that the curve passes through

$\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt[]{2}}\right)$

$\frac{3}{\sqrt[]{2}}{e}^{\alpha }-\frac{3}{\sqrt[]{2}}=e^{\alpha }+\frac{9}{2}$

$e^{\alpha }=\frac{\frac{9}{2}+\frac{3}{\sqrt[]{2}}}{\frac{3}{\sqrt[]{2}}-1}=\frac{3}{\sqrt[]{2}}\left(\frac{3+\sqrt[]{2}}{3-\sqrt[]{2}}\right)$

$f\left(x\right)=\{\begin{array}{l}x-\left[x\right]if\left[x\right]isodd\\ 1+\left[x\right]-x,if\left[x\right]iseven\end{array}$

Graph of f (x)

So

$=2{\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x\right)cos\pi xdx}$

=4

$A=\left\{\left(x,y\right):x^2\le y\le min\{x+2,4-3x\right\}$

So, area of the required region

$A={\displaystyle \underset{-1}{\overset{\frac{1}{2}1}{\int }}\left(x+2-x^2\right)dx+{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}\left(4-3x-x^2\right)}dx}$

$={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}^{\frac{1}{2}}+{\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}^1$

= $\frac{17}{6}$

$f\left(x\right)=y=a{x}^3+b{x}^2+cx+5$ ……. (i)

$\frac{dy}{dx}=3a{x}^2+2bx+c$ ……. (ii)

Touches x-axis at P (-2, 0)

$\Rightarrow y{|}_{x=-2}=0\Rightarrow -8a+4b-2c+5=0$ ……… (iii)

Touches x –axis at P (-2, 0) also implies

$\frac{dy}{dx}{|}_{x=-2}=0\Rightarrow 12a-4b+c=0$ ……… (iv)

y = f (x) cuts y-axis at (0, 5)

Given,

$\frac{dy}{dx}{|}_{x=0}=c=3$ ……. (v)

From (iii), (iv) and (v)

f (x) = 0 at x = -2 and x = 1

Local maximum value of f (x) is at x = 1

i.e., $\frac{27}{4}$

Given,

$f\left(x\right)=\underset{f_1\left(x\right)}{\underset{?}{\left(x^2-2x+7\right)}}\underset{f_2\left(x\right)}{\underset{?}{\left(e^{\left(4x^3-12x^2-180x+31\right)}\right)}}$        

f₁ (x) = x² – 2x + 7

So f (x) is decreasing in [-3, 0]

and positive also

$\therefore$ absolute maximum value of f (x) occurs at x = -3

$\therefore \alpha =-3$      

  $\Delta =\left|\begin{array}{ccc}\hfill 3sin3\theta \hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 3cos2\theta \hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right|$

= 3 sin3θ (28 – 21) + (21 cos 2θ - 18) + 1 (21 cos 2θ - 24)

$\Delta =21sin3\theta +42cos2\theta -42$          

for no solution

sin 3θ + 2 cos 2θ = 2

$\theta =\pi ,2\pi ,3\pi ,\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

$S_n=\left\{z\in C:\left|z-3+2i\right|=\frac{n}{4}\right\}$

represents a circle with centre C1 (3, 2) and radius

$r_1=\frac{n}{4}$

Similarly Tn represents circle with centre C2 (2, 3) and radius

$r_1=\frac{1}{n}$

$\sqrt[]{2}<\left|\frac{n}{4}-\frac{1}{n}\right|$

n take infinite values.